Answer
See below
Work Step by Step
a) We know $A$ is a $1\times 3$ dimension. $D$ is a $3\times 3$ dimension, then $B^T$ will be $3 \times 3$ dimension. Hence, $AD^T$ is a defined expression.
$AD^T=\begin{bmatrix}
-3 &-1 &6
\end{bmatrix}\begin{bmatrix}
-2 &0 &1\\1 &0 &-2 \\ 5&7&-1
\end{bmatrix}=\begin{bmatrix}
35 & 42& -7
\end{bmatrix}$
b) $C$ is a $2\times 4$ dimension, then $C^T$ will be $4 \times 2$ dimension. Hence, $C^TB^T$ is defined expression.
$C^TC=\begin{bmatrix}
-9 &1 \\0 & 1 \\3 &5\\-2 & -2
\end{bmatrix}\begin{bmatrix}
-9 & 0& 3 &-2\\1 & 1 & 5 &-2
\end{bmatrix}=\begin{bmatrix}
82 & 1&-22 & 16\\1 & 1 & 5 & -2\\-22 & 5 &34 &-16\\16 &-2&-16 &8
\end{bmatrix}\\
\rightarrow (C^TC)^2=\begin{bmatrix}
82 & 1&-22 & 16\\1 & 1 & 5 & -2\\-22 & 5 &34 &-16\\16 &-2&-16 &8
\end{bmatrix}\begin{bmatrix}
82 & 1&-22 & 16\\1 & 1 & 5 & -2\\-22 & 5 &34 &-16\\16 &-2&-16 &8
\end{bmatrix}=\begin{bmatrix}
7241 & -59 & -2803 & 1790\\-59 & 31 & 185 &-82\\-2803 & 185& 1921 & -1034 \\1790 &-82 &-1034 & 580
\end{bmatrix}$
c) We have $D$ is a $3\times 3$ dimension, $D^T$ will be $3 \times 3$ dimension. Since $B$ is $3\times 2$ dimension, hence $D^TB$ is a defined expression.
$D^TB=\begin{bmatrix}
-2 & 0 & 1\\1 &0&-2\\5&7&-1
\end{bmatrix}\begin{bmatrix}
0 & -4\\-7 &1\\-1 & -3
\end{bmatrix}=\begin{bmatrix}
-1 & 5\\2 & 2 \\-48 & -10
\end{bmatrix}$
